To find the speed of the ball when it was at a height of 5 m, we can use the principle of conservation of energy or the equations of motion.
Using energy conservation, the gravitational potential energy (PE) gained by the ball at its maximum height can be expressed as:
\[ PE = mgh \]
Where:
- \( m \) is the mass of the ball (which will cancel out),
- \( g \) is the acceleration due to gravity (9.8 m/s²),
- \( h \) is the height (which we will set to 25 m at its maximum height).
At its maximum height (25 m), all the kinetic energy (KE) has converted into potential energy.
When the ball is at a height of 5 m, the total mechanical energy can be expressed as the sum of the potential energy and the kinetic energy:
\[ PE = mgh = mg(5) \ KE = \frac{1}{2} mv^2 \]
At the maximum height (25 m), the potential energy is:
\[ PE_{max} = mg(25) \]
Setting the total energy at max height equal to the total energy at 5 m height, we have:
\[ mg(25) = mg(5) + \frac{1}{2} mv^2 \]
Since \( m \) appears in every term, we can cancel it out:
\[ g(25) = g(5) + \frac{1}{2} v^2 \]
Now, substituting \( g = 9.8 , \text{m/s}^2 \):
\[ 9.8(25) = 9.8(5) + \frac{1}{2} v^2 \]
Calculating the left side:
\[ 9.8 \times 25 = 245 \]
Calculating the right side:
\[ 9.8 \times 5 = 49 \]
So the equation now looks like:
\[ 245 = 49 + \frac{1}{2} v^2 \]
Subtract \( 49 \) from both sides:
\[ 245 - 49 = \frac{1}{2} v^2 \ 196 = \frac{1}{2} v^2 \]
Now, multiply both sides by 2 to solve for \( v^2 \):
\[ 392 = v^2 \]
Taking the square root to find \( v \):
\[ v = \sqrt{392} \approx 19.8 , \text{m/s} \]
Thus, the speed of the ball when it was 5 m high is approximately 19.8 m/s.
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